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Chapter 13 · Probability

Exercise 13.2 · Q17 (MCQ): Even prime number on each of a pair of dice

Exercise 13.2Independent events · Section 13.4
The question

Choose the correct answer.

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) 00    (B) 13\frac{1}{3}    (C) 112\frac{1}{12}    (D) 136\frac{1}{36}

Exam answer

The only even prime number is 2, so each die must show 2. The dice are independent.

P(2 on both dice)=16×16=136\begin{aligned} &P(2 \text{ on both dice}) \\ &= \frac{1}{6}\times\frac{1}{6} = \frac{1}{36} \end{aligned}

Answer: (D) 136\dfrac{1}{36}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

First decide which faces count. A prime has exactly two factors, and the only even one is 2 (every other even number has 2 as an extra factor). So "an even prime on each die" means the single outcome (2, 2), which has probability 16×16=136\frac16\times\frac16 = \frac{1}{36}⁠: option (D).

Given, to find, and the plan
Given
A pair of fair dice is rolled; 36 equally likely outcomes (x,y)(x, y)⁠, x,y∈{1,2,…,6}x, y \in \{1, 2, \ldots, 6\}⁠.
To find
P(an even prime number on each die), and the matching option.
Plan
(1) Find the even primes among 1 to 6. (2) Find the probability for one die. (3) Multiply, as the dice are independent; check by counting. (4) Examine every option.

Basics you need

1Prime numbers

A prime has exactly two factors, 1 and itself. On a die the primes are 2, 3 and 5; 1 is not prime.

2Classical probability

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for equally likely outcomes; two dice give 6×6=366\times 6 = 36 outcomes. Basics: P(E) = n(E)/n(S)

3Independent dice

The two dice do not affect each other, so the probability of a result on both is the product (Remark (iii), Section 13.4). Lesson: independent events

Solution

  1. Which faces are even primes?

    The faces are 1, 2, 3, 4, 5, 6. The even ones are 2, 4, 6; the primes are 2, 3, 5. The only face in both lists is 2.

    4 and 6 are even but not prime (4=2×24 = 2\times 2⁠, 6=2×36 = 2\times 3⁠), and 3 and 5 are prime but odd. In fact 2 is the only even prime of all: any larger even number has the factor 2 as well as 1 and itself.
    Definition of a prime number (earlier classes). The even primes are the faces in both lists, the intersection of the two sets. Basics: events as sets

  2. Probability for one die

    Let E1E_1 and E2E_2 be the events "the first die shows an even prime" and "the second die shows an even prime". Each means "shows 2", one face of six:

    P(E1)=P(E2)=16.P(E_1) = P(E_2) = \frac{1}{6}.
    Each face of a fair die is equally likely, and exactly one face is favourable.
    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}⁠. Basics: classical probability

  3. Multiply, because the dice are independent

    P(E1∩E2)=P(E1) P(E2)=16×16=136.\begin{aligned} P(E_1\cap E_2) &= P(E_1)\,P(E_2) \\ &= \frac{1}{6}\times\frac{1}{6} = \frac{1}{36}. \end{aligned}

    Counting gives the same: of the 36 outcomes, only (2,2)(2, 2) has an even prime on each die, so the probability is 136\frac{1}{36}⁠.

    Rolling two dice is two independent experiments, so the probability of the pair of results is the product of the separate probabilities. The count confirms it without using independence.
    Remark (iii), Section 13.4. Lesson: independent events

  4. Examine every option

    OptionValueOutcomes out of 36Correct?
    (A)000No: (2, 2) can happen
    (B)13\frac1312No: only 1 outcome is favourable
    (C)112\frac{1}{12}3No: that is 16×12\frac16\times\frac12⁠, a 2 on the first die and any even number on the second
    (D)136\frac{1}{36}1Yes
    (A) would be right only if there were no even prime, a common slip from thinking "all primes are odd". (B) and (C) correspond to 12 and 3 favourable outcomes, but the event contains only the one outcome (2, 2). (C) comes from treating "even prime" as just "even" on the second die. Only (D) matches.
    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} with n(S)=36n(S) = 36⁠. Basics: counting
Result

P(an even prime on each die) =16×16=136= \frac16\times\frac16 = \frac{1}{36}⁠, which is option (D). (Textbook answer: D.)

Check it with numbers

136≈0.0278\frac{1}{36} \approx 0.0278⁠: about 1 roll in 36. For comparison, the chance of a 2 on one die is 16≈0.1667\frac16 \approx 0.1667⁠, and 0.1667×0.1667≈0.02780.1667\times 0.1667 \approx 0.0278⁠.

Common mistakes
  1. Thinking there is no even prime. 2 is prime and even; that gives the wrong option (A).
  2. Counting 4 or 6 as primes. They have more than two factors.
  3. Using "even" for one die. The question asks for an even prime on each die.
What to take away
  1. 2 is the only even prime.
  2. For two dice, a stated result on each die has probability (one-die probability) ×\times (one-die probability).

Every fact used, and where it comes from

FactUsed inSource
2 is the only even primeStep 1Number systems (earlier classes)
P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for equally likely outcomesSteps 2, 3, 4Class XI, Probability (recalled in Section 13.1)
Independent experiments: probabilities multiplyStep 3Remark (iii), Section 13.4 (Lesson: independent events)
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